Rewrite the differential equation
Given
cos x ( 3 sin x + cos x + 3 ) d y = ( 1 + y sin x ( 3 sin x + cos x + 3 ) ) d x \cos x(3\sin x+\cos x+3)\,dy=(1+y\sin x(3\sin x+\cos x+3))\,dx cos x ( 3 sin x + cos x + 3 ) d y = ( 1 + y sin x ( 3 sin x + cos x + 3 )) d x with y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0
Divide by d x dx d x
cos x ( 3 sin x + cos x + 3 ) d y d x = 1 + y sin x ( 3 sin x + cos x + 3 ) \cos x(3\sin x+\cos x+3)\frac{dy}{dx}=1+y\sin x(3\sin x+\cos x+3) cos x ( 3 sin x + cos x + 3 ) d x d y = 1 + y sin x ( 3 sin x + cos x + 3 ) So,
d y d x = 1 cos x ( 3 sin x + cos x + 3 ) + y tan x \frac{dy}{dx}=\frac{1}{\cos x(3\sin x+\cos x+3)}+y\tan x d x d y = cos x ( 3 sin x + cos x + 3 ) 1 + y tan x Hence
d y d x − y tan x = 1 cos x ( 3 sin x + cos x + 3 ) \frac{dy}{dx}-y\tan x=\frac{1}{\cos x(3\sin x+\cos x+3)} d x d y − y tan x = cos x ( 3 sin x + cos x + 3 ) 1 This is a linear differential equation.
Find the integrating factor
For
d y d x + P ( x ) y = Q ( x ) , \frac{dy}{dx}+P(x)y=Q(x), d x d y + P ( x ) y = Q ( x ) , we have
P ( x ) = − tan x P(x)=-\tan x P ( x ) = − tan x Therefore the integrating factor is
I.F. = e ∫ − tan x d x = e ln ( cos x ) = cos x \text{I.F.}=e^{\int -\tan x\,dx}=e^{\ln(\cos x)}=\cos x I.F. = e ∫ − t a n x d x = e l n ( c o s x ) = cos x (since 0 ≤ x ≤ π / 2 0\le x\le \pi/2 0 ≤ x ≤ π /2 cos x ≥ 0 \cos x\ge 0 cos x ≥ 0
Multiplying the equation by cos x \cos x cos x
cos x d y d x − y sin x = 1 3 sin x + cos x + 3 \cos x\frac{dy}{dx}-y\sin x=\frac{1}{3\sin x+\cos x+3} cos x d x d y − y sin x = 3 sin x + cos x + 3 1 But
d d x ( y cos x ) = cos x d y d x − y sin x \frac{d}{dx}(y\cos x)=\cos x\frac{dy}{dx}-y\sin x d x d ( y cos x ) = cos x d x d y − y sin x Thus,
d d x ( y cos x ) = 1 3 sin x + cos x + 3 \frac{d}{dx}(y\cos x)=\frac{1}{3\sin x+\cos x+3} d x d ( y cos x ) = 3 sin x + cos x + 3 1
Integrate
Integrate from 0 0 0 x x x
y cos x − y ( 0 ) cos 0 = ∫ 0 x d t 3 sin t + cos t + 3 y\cos x- y(0)\cos 0=\int_0^x \frac{dt}{3\sin t+\cos t+3} y cos x − y ( 0 ) cos 0 = ∫ 0 x 3 sin t + cos t + 3 d t Since y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0
y cos x = ∫ 0 x d t 3 sin t + cos t + 3 y\cos x=\int_0^x \frac{dt}{3\sin t+\cos t+3} y cos x = ∫ 0 x 3 sin t + cos t + 3 d t So at x = π / 3 x=\pi/3 x = π /3
y ( π 3 ) cos π 3 = ∫ 0 π / 3 d x 3 sin x + cos x + 3 y\left(\frac\pi3\right)\cos\frac\pi3=\int_0^{\pi/3}\frac{dx}{3\sin x+\cos x+3} y ( 3 π ) cos 3 π = ∫ 0 π /3 3 sin x + cos x + 3 d x Since cos ( π / 3 ) = 1 / 2 \cos(\pi/3)=1/2 cos ( π /3 ) = 1/2
y ( π 3 ) = 2 ∫ 0 π / 3 d x 3 sin x + cos x + 3 y\left(\frac\pi3\right)=2\int_0^{\pi/3}\frac{dx}{3\sin x+\cos x+3} y ( 3 π ) = 2 ∫ 0 π /3 3 sin x + cos x + 3 d x
Evaluate the integral using t = tan x 2 t=\tan\frac x2 t = tan 2 x
Use
sin x = 2 t 1 + t 2 , cos x = 1 − t 2 1 + t 2 , d x = 2 d t 1 + t 2 \sin x=\frac{2t}{1+t^2},\qquad \cos x=\frac{1-t^2}{1+t^2},\qquad dx=\frac{2\,dt}{1+t^2} sin x = 1 + t 2 2 t , cos x = 1 + t 2 1 − t 2 , d x = 1 + t 2 2 d t Then
3 sin x + cos x + 3 = 6 t + 1 − t 2 1 + t 2 + 3 = 6 t + 1 − t 2 + 3 + 3 t 2 1 + t 2 = 2 ( t 2 + 3 t + 2 ) 1 + t 2 3\sin x+\cos x+3
=\frac{6t+1-t^2}{1+t^2}+3
=\frac{6t+1-t^2+3+3t^2}{1+t^2}
=\frac{2(t^2+3t+2)}{1+t^2} 3 sin x + cos x + 3 = 1 + t 2 6 t + 1 − t 2 + 3 = 1 + t 2 6 t + 1 − t 2 + 3 + 3 t 2 = 1 + t 2 2 ( t 2 + 3 t + 2 ) Thus
d x 3 sin x + cos x + 3 = 2 d t 1 + t 2 2 ( t 2 + 3 t + 2 ) 1 + t 2 = d t t 2 + 3 t + 2 = d t ( t + 1 ) ( t + 2 ) \frac{dx}{3\sin x+\cos x+3}
=\frac{\frac{2dt}{1+t^2}}{\frac{2(t^2+3t+2)}{1+t^2}}
=\frac{dt}{t^2+3t+2}
=\frac{dt}{(t+1)(t+2)} 3 sin x + cos x + 3 d x = 1 + t 2 2 ( t 2 + 3 t + 2 ) 1 + t 2 2 d t = t 2 + 3 t + 2 d t = ( t + 1 ) ( t + 2 ) d t Now the limits:
when x = 0 x=0 x = 0 t = tan 0 = 0 t=\tan 0=0 t = tan 0 = 0
when x = π / 3 x=\pi/3 x = π /3 t = tan ( π / 6 ) = 1 3 t=\tan(\pi/6)=\frac1{\sqrt3} t = tan ( π /6 ) = 3 1
So
∫ 0 π / 3 d x 3 sin x + cos x + 3 = ∫ 0 1 / 3 d t ( t + 1 ) ( t + 2 ) \int_0^{\pi/3}\frac{dx}{3\sin x+\cos x+3}
=\int_0^{1/\sqrt3}\frac{dt}{(t+1)(t+2)} ∫ 0 π /3 3 sin x + cos x + 3 d x = ∫ 0 1/ 3 ( t + 1 ) ( t + 2 ) d t Use partial fractions:
1 ( t + 1 ) ( t + 2 ) = 1 t + 1 − 1 t + 2 \frac{1}{(t+1)(t+2)}=\frac{1}{t+1}-\frac{1}{t+2} ( t + 1 ) ( t + 2 ) 1 = t + 1 1 − t + 2 1 Hence
∫ 0 1 / 3 d t ( t + 1 ) ( t + 2 ) = [ ln ( t + 1 ) − ln ( t + 2 ) ] 0 1 / 3 \int_0^{1/\sqrt3}\frac{dt}{(t+1)(t+2)}
=\left[\ln(t+1)-\ln(t+2)\right]_0^{1/\sqrt3} ∫ 0 1/ 3 ( t + 1 ) ( t + 2 ) d t = [ ln ( t + 1 ) − ln ( t + 2 ) ] 0 1/ 3 = ln ( 1 + 1 / 3 2 + 1 / 3 ) − ln ( 1 2 ) =\ln\left(\frac{1+1/\sqrt3}{2+1/\sqrt3}\right)-\ln\left(\frac12\right) = ln ( 2 + 1/ 3 1 + 1/ 3 ) − ln ( 2 1 ) = ln ( 2 ( 1 + 1 / 3 ) 2 + 1 / 3 ) =\ln\left(\frac{2(1+1/\sqrt3)}{2+1/\sqrt3}\right) = ln ( 2 + 1/ 3 2 ( 1 + 1/ 3 ) ) Multiply numerator and denominator by 3 \sqrt3 3
= ln ( 2 ( 3 + 1 ) 2 3 + 1 ) =\ln\left(\frac{2(\sqrt3+1)}{2\sqrt3+1}\right) = ln ( 2 3 + 1 2 ( 3 + 1 ) ) Therefore,
y ( π 3 ) = 2 ln ( 2 ( 3 + 1 ) 2 3 + 1 ) y\left(\frac\pi3\right)=2\ln\left(\frac{2(\sqrt3+1)}{2\sqrt3+1}\right) y ( 3 π ) = 2 ln ( 2 3 + 1 2 ( 3 + 1 ) ) Now simplify:
2 ( 3 + 1 ) 2 3 + 1 \frac{2(\sqrt3+1)}{2\sqrt3+1} 2 3 + 1 2 ( 3 + 1 ) Multiply numerator and denominator by ( 2 3 − 1 ) (2\sqrt3-1) ( 2 3 − 1 )
2 ( 3 + 1 ) ( 2 3 − 1 ) ( 2 3 + 1 ) ( 2 3 − 1 ) = 2 ( 6 − 3 + 2 3 − 1 ) 12 − 1 = 2 ( 5 + 3 ) 11 = 10 + 2 3 11 \frac{2(\sqrt3+1)(2\sqrt3-1)}{(2\sqrt3+1)(2\sqrt3-1)}
=\frac{2(6-\sqrt3+2\sqrt3-1)}{12-1}
=\frac{2(5+\sqrt3)}{11}
=\frac{10+2\sqrt3}{11} ( 2 3 + 1 ) ( 2 3 − 1 ) 2 ( 3 + 1 ) ( 2 3 − 1 ) = 12 − 1 2 ( 6 − 3 + 2 3 − 1 ) = 11 2 ( 5 + 3 ) = 11 10 + 2 3 Hence
y ( π 3 ) = 2 ln ( 2 3 + 10 11 ) y\left(\frac\pi3\right)=2\ln\left(\frac{2\sqrt3+10}{11}\right) y ( 3 π ) = 2 ln ( 11 2 3 + 10 )
Compare with options
This matches Option C :
2 log e ( 2 3 + 10 11 ) 2\log_e\left(\frac{2\sqrt3+10}{11}\right) 2 log e ( 11 2 3 + 10 )
Comparison with stored correct answer
Stored correct answer is C , which matches the derived result.